求助一道最值问题
求助，想了很长时间也没什么头绪 利用60度角对应的△ABC,△ADE,△ADF，三个余弦定理理论上可以解出 Clear["Global`*"];(*计算余弦值子函数,利用三边计算余弦值*)
cs:=(a^2+b^2-c^2)/(2*a*b)
FullSimplify@Maximize[
{DF,
(*假设BD=EC=FC=x,则BC=4*x*)
1/2==cs&&
1/2==cs&&
1/2==cs&&
AD>0&&
AE>0&&
DF>0&&
x>0
},{AD,AE,DF,x}
]
N[%]
三次利用余弦定理
求解结果如下：
\[\left\{\frac{2}{15} \left(\sqrt{259}+8\right),\left\{\text{AD}\to \sqrt{\frac{76}{307}+\frac{816}{307 \sqrt{259}}},\text{AE}\to \sqrt{\frac{1292}{307}+\frac{4048}{307 \sqrt{259}}},\text{DF}\to \frac{2}{15} \left(\sqrt{259}+8\right),x\to \frac{2}{15} \sqrt{17+\frac{76}{\sqrt{259}}}\right\}\right\}\]
数值如下
{3.21246,{AD->0.64243,AE->2.24227,DF->3.21246,x->0.621431}}
你缺的不是头绪，你缺的是计算软件！ Clear["Global`*"];
(*计算余弦值子函数,角度60*)
fun:=a^2+b^2-2*a*b*(1/2)-c^2
(*三个约数条件*)
f1=fun
f2=fun
f3=fun
(*拉格朗日乘子法目标函数+约数条件*)
yy=DF+x1*f1+x2*f2+x3*f3
(*求解七个偏导数*)
yyAD=D
yyAE=D
yyDF=D
yyx=D
yyx1=D
yyx2=D
yyx3=D
(*联立求解方程组*)
out=Solve[
yyAD==0&&
yyAE==0&&
yyDF==0&&
yyx==0&&
yyx1==0&&
yyx2==0&&
yyx3==0,
{AD,AE,DF,x,x1,x2,x3}
];
out1=FullSimplify
(*转化成根式显式表达*)
Grid@ToRadicals
N[%]
求解结果
\[\begin{array}{ccccccc}
\text{AD}\to -\sqrt{\frac{76}{307}+\frac{816}{307 \sqrt{259}}} & \text{AE}\to -2 \sqrt{\frac{323}{307}+\frac{1012}{307 \sqrt{259}}} & \text{DF}\to \frac{1}{15} (-2) \left(\sqrt{259}+8\right) & x\to \frac{1}{15} (-2) \sqrt{17+\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(\sqrt{259}+8\right) & \text{x2}\to -\frac{1}{8} & \text{x3}\to \frac{1}{52} \left(8-\sqrt{259}\right) \\
\text{AD}\to -\sqrt{\frac{76}{307}+\frac{816}{307 \sqrt{259}}} & \text{AE}\to -2 \sqrt{\frac{323}{307}+\frac{1012}{307 \sqrt{259}}} & \text{DF}\to \frac{2}{15} \left(\sqrt{259}+8\right) & x\to \frac{1}{15} (-2) \sqrt{17+\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(-\sqrt{259}-8\right) & \text{x2}\to \frac{1}{8} & \text{x3}\to \frac{1}{52} \left(\sqrt{259}-8\right) \\
\text{AD}\to \sqrt{\frac{76}{307}+\frac{816}{307 \sqrt{259}}} & \text{AE}\to \sqrt{\frac{1292}{307}+\frac{4048}{307 \sqrt{259}}} & \text{DF}\to \frac{1}{15} (-2) \left(\sqrt{259}+8\right) & x\to \frac{2}{15} \sqrt{17+\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(\sqrt{259}+8\right) & \text{x2}\to -\frac{1}{8} & \text{x3}\to \frac{1}{52} \left(8-\sqrt{259}\right) \\
\text{AD}\to \sqrt{\frac{76}{307}+\frac{816}{307 \sqrt{259}}} & \text{AE}\to \sqrt{\frac{1292}{307}+\frac{4048}{307 \sqrt{259}}} & \text{DF}\to \frac{2}{15} \left(\sqrt{259}+8\right) & x\to \frac{2}{15} \sqrt{17+\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(-\sqrt{259}-8\right) & \text{x2}\to \frac{1}{8} & \text{x3}\to \frac{1}{52} \left(\sqrt{259}-8\right) \\
\text{AD}\to -2 \sqrt{\frac{4921-204 \sqrt{259}}{79513}} & \text{AE}\to 2 \sqrt{\frac{323}{307}-\frac{1012}{307 \sqrt{259}}} & \text{DF}\to \frac{1}{15} (-2) \left(\sqrt{259}-8\right) & x\to \frac{1}{15} (-2) \sqrt{17-\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(\sqrt{259}-8\right) & \text{x2}\to \frac{1}{8} & \text{x3}\to \frac{1}{52} \left(-\sqrt{259}-8\right) \\
\text{AD}\to -2 \sqrt{\frac{4921-204 \sqrt{259}}{79513}} & \text{AE}\to 2 \sqrt{\frac{323}{307}-\frac{1012}{307 \sqrt{259}}} & \text{DF}\to \frac{2}{15} \left(\sqrt{259}-8\right) & x\to \frac{1}{15} (-2) \sqrt{17-\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(8-\sqrt{259}\right) & \text{x2}\to -\frac{1}{8} & \text{x3}\to \frac{1}{52} \left(\sqrt{259}+8\right) \\
\text{AD}\to 2 \sqrt{\frac{4921-204 \sqrt{259}}{79513}} & \text{AE}\to -2 \sqrt{\frac{323}{307}-\frac{1012}{307 \sqrt{259}}} & \text{DF}\to \frac{1}{15} (-2) \left(\sqrt{259}-8\right) & x\to \frac{2}{15} \sqrt{17-\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(\sqrt{259}-8\right) & \text{x2}\to \frac{1}{8} & \text{x3}\to \frac{1}{52} \left(-\sqrt{259}-8\right) \\
\text{AD}\to 2 \sqrt{\frac{4921-204 \sqrt{259}}{79513}} & \text{AE}\to -2 \sqrt{\frac{323}{307}-\frac{1012}{307 \sqrt{259}}} & \text{DF}\to \frac{2}{15} \left(\sqrt{259}-8\right) & x\to \frac{2}{15} \sqrt{17-\frac{76}{\sqrt{259}}} & \text{x1}\to \frac{1}{60} \left(8-\sqrt{259}\right) & \text{x2}\to -\frac{1}{8} & \text{x3}\to \frac{1}{52} \left(\sqrt{259}+8\right) \\
\end{array}\]
\[\begin{array}{ccccccc}
\text{AD}\to -0.64243 & \text{AE}\to -2.24227 & \text{DF}\to -3.21246 & x\to -0.621431 & \text{x1}\to 0.401558 & \text{x2}\to -0.125 & \text{x3}\to -0.155644 \\
\text{AD}\to -0.64243 & \text{AE}\to -2.24227 & \text{DF}\to 3.21246 & x\to -0.621431 & \text{x1}\to -0.401558 & \text{x2}\to 0.125 & \text{x3}\to 0.155644 \\
\text{AD}\to 0.64243 & \text{AE}\to 2.24227 & \text{DF}\to -3.21246 & x\to 0.621431 & \text{x1}\to 0.401558 & \text{x2}\to -0.125 & \text{x3}\to -0.155644 \\
\text{AD}\to 0.64243 & \text{AE}\to 2.24227 & \text{DF}\to 3.21246 & x\to 0.621431 & \text{x1}\to -0.401558 & \text{x2}\to 0.125 & \text{x3}\to 0.155644 \\
\text{AD}\to -0.287051 & \text{AE}\to 1.84096 & \text{DF}\to -1.07913 & x\to -0.467192 & \text{x1}\to 0.134891 & \text{x2}\to 0.125 & \text{x3}\to -0.463336 \\
\text{AD}\to -0.287051 & \text{AE}\to 1.84096 & \text{DF}\to 1.07913 & x\to -0.467192 & \text{x1}\to -0.134891 & \text{x2}\to -0.125 & \text{x3}\to 0.463336 \\
\text{AD}\to 0.287051 & \text{AE}\to -1.84096 & \text{DF}\to -1.07913 & x\to 0.467192 & \text{x1}\to 0.134891 & \text{x2}\to 0.125 & \text{x3}\to -0.463336 \\
\text{AD}\to 0.287051 & \text{AE}\to -1.84096 & \text{DF}\to 1.07913 & x\to 0.467192 & \text{x1}\to -0.134891 & \text{x2}\to -0.125 & \text{x3}\to 0.463336 \\
\end{array}\] 谢谢大家，现在找到答案了，属于阿氏圆最值问题 .Sun 发表于 2020-3-15 17:26
谢谢大家，现在找到答案了，属于阿氏圆最值问题
奇技淫巧！
代数、微积分的办法，多么自然而然的思路呀！ 本帖最后由 王守恩 于 2020-3-17 09:21 编辑
谢谢 mathematica！依样画葫芦：FullSimplify@Maximize
换已知条件 DE=2 为 DE=sin60°，记 DA=sin(x)，DB=sin(y)
可得
\((\sin(x)+\sin(y))^2+(\sin(120^\circ-x)+\sin(y))^2-2(\sin(x)+\sin(y))(\sin(120^\circ-x)+\sin(y))\cos(60^\circ)=(4\sin(y))^2\)
求\(\ \ \sqrt{(\sin(60^\circ))^2+(2\sin(y))^2+2\sin(60^\circ)(2\sin(y))\cos(x)}\ \ \)的最大值
化简
\(30\cos(2y)+2\sqrt{3}\cos(x)\sin(y)+6\sin(x)\sin(y)=27\)
求\(\ \ \sqrt{\frac{3}{4}+2\sqrt{3}\cos(x)\sin(y)+(2\sin(y))^2}\ \ \)的最大值 王守恩 发表于 2020-3-17 08:38
谢谢 mathematica！依样画葫芦：FullSimplify@Maximize
换已知条件 DE=2 为 DE=sin60°，记 DA=sin(x) ...
对你的不感兴趣！
我只对有推理过程的感兴趣，
其余乱猜测胡乱说的，不感兴趣
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